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  1. Prepare
  2. Algorithms
  3. Strings
  4. Mars Exploration
  5. Discussions

Mars Exploration

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1066 Discussions

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  • calancea_bianca1
     + 0 comments
    def marsExploration(s):
    # Write your code here
    count=0
    for i in range(1,len(s)-1,3):
        if s[i-1]!='S':
            count+=1
        if s[i]!='O':
            count+=1
        if s[i+1]!='S':
            count+=1
    return count

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can find the explanation here : https://youtu.be/Gf0fpGE83r4

    int marsExploration(string s) {
   int result = 0;
    string base = "SOS";
    for(int i = 0; i < s.size(); i++) if(s[i] != base[i%3]) result++;
    return result;
}

  • patrickgao2020
     + 0 comments

    Here is my one line Python solution!

    def marsExploration(s):
    return len([True for letter in range(len(s)) if s[letter] != ["S", "O", "S"][letter % 3]])

  • highsoft_soi
     + 0 comments

    Perl solution:

    sub marsExploration {
    my $s = shift;
    
    my @arr = split("", $s);
    my $cnt = 0;
    for (my $i = 0; $i <= scalar(@arr) - 1; $i += 3) {
        $cnt++ if ($arr[$i] ne "S");
        $cnt++ if ($arr[$i+1] ne "O");
        $cnt++ if ($arr[$i+2] ne "S");
    }
    
    return $cnt
}

  • vuhuutuan0
     + 0 comments

    My answer in Typescript, simple, not minimized

    const SOS = (l: number) => 'SOS'.repeat(l);
const SBD = (l: number, cb: (i: number) => string) => Array(l).fill('').map((_, i) => cb(i)).join('');
function marsExploration(s: string): number {
    let recieved_signal = s;
    let expected_signal = SOS(s.length / 3);
    let differen_signal = SBD(s.length, i => recieved_signal[i] != expected_signal[i] ? 'X' : ' ');

    console.log('Expected signal:', expected_signal)
    console.log('Recieved signal:', recieved_signal)
    console.log('Difference:     ', differen_signal)

    return differen_signal.split('').reduce((p, c) => c == 'X' ? p + 1 : p, 0)
}